The number of four-digit numbers that can be | vedclass

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(D) Step $1$: Choose $2$ distinct digits out of $9$ available digits ($1$ to $9$). This can be done in $^9C_2 = \frac{9 	imes 8}{2} = 36$ ways.Step $

Vedclass · Accepted Answer

$504$

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(D) Step $1$: Choose $2$ distinct digits out of $9$ available digits ($1$ to $9$). This can be done in $^9C_2 = \frac{9 	imes 8}{2} = 36$ ways.Step $2$: For each pair of chosen digits,we need to form a $4$-digit number using both digits at least once.Step $3$: The total number of ways to fill $4$ positions using $2$ chosen digits is $2^4 = 16$.Step $4$: We must exclude the cases where only one digit is used (i.e.,all $4$ digits are the same). There are $2$ such cases (all $4$ digits are the first chosen digit or all $4$ digits are the second chosen digit).Step $5$: The number of valid $4$-digit numbers for each pair is $(2^4 - 2) = 14$.Step $6$: Total numbers = $36 	imes 14 = 504$.

The number of four-digit numbers that can be formed using all the digits from $1$ to $9$ (excluding zero) such that every number has exactly $2$ distinct digits is:

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